Question

If A =1i/√2 + bj is a unit vector, then value of b:​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\vec{A}=\dfrac{1}{\sqrt{2}}\,\hat{i}+b\hat{j}}$

Since the given vector is a unit vector, so,

$\sf{|\vec{A}|=1}$

$\sf{\implies\,\sqrt{\bigg(\dfrac{1}{\sqrt{2}}\bigg)^2+(b)^2}=1}$

$\sf{\implies\,\dfrac{1}{2}+b^2=1}$

$\sf{\implies\,b^2=1-\dfrac{1}{2}}$

$\sf{\implies\,b^2=\dfrac{1}{2}}$

$\sf{\implies\,b=\pm\dfrac{1}{\sqrt{2}}}$